Your final statement is not quite accurate, because there is an element $a$ with $a*a=e$ that is not of order $2$, to wit $e$ itself. In fact, in the situation you describe, the number of elements $a$ of $G$ that satisfy $aa=e$ will be even, because there will be an odd number of elements of order $2$, and you'll also have the identity element.
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Arturo MagidinMay 29 '11 at 22:55

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@Arturo: One might use this to generalize the result to say that the number of square roots of the identity element always has the same parity as the order of the group.
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jorikiMay 29 '11 at 23:25

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@joriki: Indeed. Of course, even more is true: Frobenius's Theorem gives that the number of solutions to $x^n=1$ in a finite group $G$ is always a multiple of $\gcd(n,|G|)$. For $n=2$, the theorem implies the case of $|G|$ even, while Lagrange implies the case of $|G|$ odd.
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Arturo MagidinMay 30 '11 at 2:00

Thank you Arturo. This helped to make the problem very clear.
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JacobsenMay 29 '11 at 23:23

@Arturo - i remember encountering this problem a few months ago as an exercise in prof Milne's notes on group theory chapter 1 - which is as far as i got - had me stumped until i twigged the obvious. your answer is a very elegant way of expressing the matter. as yet i have not learned how to vote, but i would certainly give it a plus if i knew how.
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David HoldenDec 11 '13 at 7:57

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PS might one also express the same idea in terms of the kernel of the anti-isomorphism $g \rightarrow g^{-1}$, or is there a taboo on using this naughty mapping?
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David HoldenDec 11 '13 at 8:14

@DavidHolden As one doesn't need anti-isomorphism, it suffices to note this is an involution and count fixpoints
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Hagen von EitzenJun 20 at 8:57